Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

*12(s1(x), y) -> +12(*2(x, y), y)
*12(s1(x), y) -> *12(x, y)
*12(p1(x), y) -> *12(x, y)
*12(p1(x), y) -> +12(*2(x, y), minus1(y))
MINUS1(p1(x)) -> MINUS1(x)
MINUS1(s1(x)) -> MINUS1(x)
+12(s1(x), y) -> +12(x, y)
+12(p1(x), y) -> +12(x, y)
*12(p1(x), y) -> MINUS1(y)

The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

*12(s1(x), y) -> +12(*2(x, y), y)
*12(s1(x), y) -> *12(x, y)
*12(p1(x), y) -> *12(x, y)
*12(p1(x), y) -> +12(*2(x, y), minus1(y))
MINUS1(p1(x)) -> MINUS1(x)
MINUS1(s1(x)) -> MINUS1(x)
+12(s1(x), y) -> +12(x, y)
+12(p1(x), y) -> +12(x, y)
*12(p1(x), y) -> MINUS1(y)

The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

MINUS1(s1(x)) -> MINUS1(x)
MINUS1(p1(x)) -> MINUS1(x)

The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MINUS1(s1(x)) -> MINUS1(x)
MINUS1(p1(x)) -> MINUS1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MINUS1(x1)) = 2·x1   
POL(p1(x1)) = 2 + 2·x1   
POL(s1(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

+12(s1(x), y) -> +12(x, y)
+12(p1(x), y) -> +12(x, y)

The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


+12(s1(x), y) -> +12(x, y)
+12(p1(x), y) -> +12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+12(x1, x2)) = 2·x1   
POL(p1(x1)) = 2 + 2·x1   
POL(s1(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

*12(s1(x), y) -> *12(x, y)
*12(p1(x), y) -> *12(x, y)

The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


*12(s1(x), y) -> *12(x, y)
*12(p1(x), y) -> *12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(*12(x1, x2)) = 2·x1   
POL(p1(x1)) = 2 + 2·x1   
POL(s1(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.